[proofplan] We prove the identity by induction on the positive integer $m$. The base case is exactly the eigenvector equation $Av=\lambda v$. For the induction step, we multiply the induction hypothesis by $A$ and then use scalar compatibility of matrix multiplication together with the eigenvector equation. [/proofplan] [step:Verify the identity for the first power] Since $v$ is an eigenvector of $A$ with eigenvalue $\lambda$, by definition $v \ne 0$ and \begin{align*} Av = \lambda v. \end{align*} Because $A^1=A$ and $\lambda^1=\lambda$, this is precisely \begin{align*} A^1v=\lambda^1v. \end{align*} Thus the assertion holds for $m=1$. [/step] [step:Propagate the formula from one power to the next] Let $r \in \mathbb{N}$, and assume as the induction hypothesis that \begin{align*} A^r v = \lambda^r v. \end{align*} Using the recursive definition of matrix powers, associativity of matrix multiplication, the induction hypothesis, compatibility of scalar multiplication with the [linear map](/page/Linear%20Map) represented by $A$, and the eigenvector equation, we obtain \begin{align*} A^{r+1}v = A(A^r v). \end{align*} Substituting the induction hypothesis gives \begin{align*} A(A^r v)=A(\lambda^r v). \end{align*} Since $\lambda^r \in k$ is a scalar and $A:k^n\to k^n$ is $k$-linear, this becomes \begin{align*} A(\lambda^r v)=\lambda^r Av. \end{align*} Using $Av=\lambda v$, we get \begin{align*} \lambda^r Av=\lambda^r(\lambda v). \end{align*} By associativity of multiplication in the field $k$ and scalar multiplication on $k^n$, \begin{align*} \lambda^r(\lambda v)=\lambda^{r+1}v. \end{align*} Therefore \begin{align*} A^{r+1}v=\lambda^{r+1}v. \end{align*} [guided] Let $r \in \mathbb{N}$, and suppose the desired formula is already known at the exponent $r$: \begin{align*} A^r v = \lambda^r v. \end{align*} We want to prove the same formula with exponent $r+1$. The reason to multiply by $A$ is that the recursive definition of matrix powers expresses the next power as \begin{align*} A^{r+1}=AA^r. \end{align*} Applying both sides to the vector $v$ gives \begin{align*} A^{r+1}v=A(A^r v). \end{align*} Now we insert the induction hypothesis into the vector being acted on by $A$: \begin{align*} A(A^r v)=A(\lambda^r v). \end{align*} Here $\lambda^r$ is a scalar in $k$, and the matrix $A$ represents a $k$-linear map $A:k^n\to k^n$. Therefore scalar factors may be pulled through the action of $A$: \begin{align*} A(\lambda^r v)=\lambda^r Av. \end{align*} The remaining input is exactly the eigenvector equation. Since $v$ is an eigenvector of $A$ with eigenvalue $\lambda$, we have \begin{align*} Av=\lambda v. \end{align*} Substituting this into the previous expression yields \begin{align*} \lambda^r Av=\lambda^r(\lambda v). \end{align*} Finally, scalar multiplication on $k^n$ is associative with multiplication in $k$, and the recursive definition of scalar powers gives $\lambda^r\lambda=\lambda^{r+1}$. Hence \begin{align*} \lambda^r(\lambda v)=\lambda^{r+1}v. \end{align*} Combining the displayed equalities gives \begin{align*} A^{r+1}v=\lambda^{r+1}v. \end{align*} [/guided] [/step] [step:Conclude by induction] The base case $m=1$ has been proved, and the preceding step shows that validity at an arbitrary positive integer $r$ implies validity at $r+1$. By induction on positive integers, we obtain \begin{align*} A^m v=\lambda^m v \end{align*} for every positive integer $m$. This is the desired statement. [/step]